Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The expectation value of the electron’s kinetic energy in the hydrogen ground state equals the magnitude of the total energy (see Exercise 60). What must be the width of a cubic finite wall, in terms of a0, for this ground state to have this same energy?

Short Answer

Expert verified

The width of the cubic finite wall should be 0.29 nm, about five times the Bohr radii.

Step by step solution

01

A concept:

Energy of the cubic finite wall can be given by,

E1,1,1=3π2h22mL2

Where, his Planck’s constant, Lis the width of the wall, and mis the mass.

02

Width of the cubic finite wall:

Rewrite the equation for energy as below.
13.6eV=3π2h22mL2L=πh32m×13.6eV=3.14(1.055×10-34J.s)2(9.11×10-31kg(13.6×1.6×1019J)/3=0.29nm

Hence, width of the wall is 0.29 nm .

03

Width in terms of a0:

As you know that,a0=0.05nm

Where,a0 is the radius of the hydrogen atom.

Hence, the obtained width of the wall is about five times that of the a0.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,λ=2L/n , where is the length of its home. Given that λ=h/p, we would have p=nh/2L, and the kinetic energy, p2/2m, would thus be n2h2/8mL2. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: E=U. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?

Taking then=3states as representative, explain the relationship between the complexity numbers of nodes and antinodes-of hydrogen's standing waves in the radial direction and their complexity in the angular direction at a given value of n. Is it a direct or inverse relationship, and why?

Here we Pursue the more rigorous approach to the claim that the property quantized according to ml is Lz,

(a) Starting with a straightforward application of the chain rule,

φ=xφ/x+yφy+zφz

Use the transformations given in Table 7.2 to show that

φ=-yx+xy

(b) Recall that L = r x p. From the z-component of this famous formula and the definition of operators for px and py, argue that the operator for Lz is -ihφ..

(c) What now allows us to say that our azimuthal solutioneimlφ has a well-defined z-component of angular momentum and that is value mlh.

A particle orbiting due to an attractive central force has angular momentum L=1.00×10-33kg.m/s What z-components of angular momentum is it possible to detect?

Which electron transitions in singly ionized helium yield photon in the 450 - 500 nm(blue) portion of the visible range, and what are their wavelengths?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free